# How to prove that $\lim_{k\rightarrow\infty} H_k-\log_b(x)$ converges/diverges?

It is known that: $$\lim_{k\rightarrow\infty} H_k-\log(k)=\gamma$$Where $$\gamma$$ is Euler's constant. What if we change the base from $$e$$ to any other number $$b$$? How do we prove that this converges/diverges? I realized that for $$b=3$$ it seems to converge but then starts to grow slowly.

• 1) Surely you mean $\log(k)$ instead of $\log(x)$. 2) Note that $\log_b(k) = \frac{\log k}{\log b}$. So if $b\neq e,$ the difference $H_k - \log_b(k)$ must diverge - for large $k$, this is $\gamma + o(1) + (1-1/\log(b))\cdot \log k$ which blows up with $k$. Dec 25, 2022 at 23:19

Note that $$\log(k)= \int\limits_1^k\frac{1}{x}\mathrm{d}x$$, so $$H_{k-1}$$ is a upper sum of that Integral and $$\sum\limits_{n=2}^{k}\frac{1}{n}=H_k-1$$ is a lower sum. Therefore we can bound $$H_k-\log{(k)}$$ via $$H_k-H_{k-1}\le H_k-\log{(k)}\le H_k-(H_k-1) \Leftrightarrow \frac{1}{k}\le H_k-\log(k)\le 1$$.
Edit: I kinda misread the question, for different bases, the limit will diverge because, as pointed out in the comments $$H_k-\log_b(k)=-\log_b(k)+\log(k)+\gamma + \frac{1}{2k}+o(\frac{1}{k^2})$$, which diverges